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1 gram of commercial AgNO3 is dissolved in 50 ml water.It is treated with 50 ml of KI solution.The silver iodide thus precipitated is filtered off.The excess KI in the filtrate is titrated with M/10 KIO3 solution in presence of 6N HCl till all the I- is converted to ICl.It requires 50 ml of M/10 KIO3.Under similar conditions 20 ml of the same stock solution of KI required 30 ml of M/10 KIO3.Calculate the % of siver nitrate in the sample.
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6 Answers
1357KIO3+2KI+6HCl→3ICl+3KCl+3H2O
mmol of KIO3 reacted with KI=50*0.1=5
mmol 0f KI reacted with KIO3=2*5=10
in another sample mmol of KI used(in 20 ml)=6
if volume is 50 ml then mmol of KI =6*2.5=15
mmol of KI reacted with AgNO3=15-10=5
mmol of AgNO3=5
wt. of AgNO3=5*170/1000=0.85 gm
wt%=(0.85/1)*100=85%
24meq KI in 20ml=meq KIO3=30*4/10=12
meq KI in 50ml=12*50/20=30
meq KI left after reaction with AgNO#=50*4/10=20
=> meq used by AgNO3=10
n factor KI=2
=>meq KI used by AgNO3 for n=1 => 5
meq AgNO3=5 =>w/170 *1000=0.85g
=> %=85%
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